Write down the network as data
N: {S→A:5, S→B:6, A→C:4, A→D:3, B→D:5, C→T:6, D→T:4, B→C:2} The source, the sink, the arcs and their capacities describe the whole network.
Write down the given flow
f: {S→A:4, S→B:4, A→C:4, A→D:0, B→D:4, C→T:4, D→T:4, B→C:0} Each arc carries the amount of flow shown.
Add the flow along the arcs leaving the source S
val(f)=4+4=8 The value of a flow is the net flow out of the source.
Recall the value of a flow
val(f)=v∑f(S,v)−v∑f(v,S) The value of a flow is the net flow out of the source.
Recall flow conservation at an intermediate vertex
u∑f(u,x)=w∑f(x,w) At every vertex except the source and the sink, flow in equals flow out.
Recall the capacity constraint on each arc
0≤f(u,v)≤c(u,v) No arc may carry more flow than its capacity.
Recall the definition of a cut
a cut splits the vertices into a source side and a sink side The source lies on one side of the cut and the sink on the other.
Recall the capacity of a cut
C=u∈S-side, v∈T-side∑c(u,v) Only arcs crossing from the source side to the sink side are counted.
Recall the maximum-flow minimum-cut theorem
maxval(f)=minC The value of a maximum flow equals the capacity of a minimum cut.
Recall the weak-duality bound
val(f)≤C Every flow value is at most the capacity of every cut.
Recall how a cut proves optimality
val(f)=C ⇒ f is maximal If a flow and a cut have the same size, neither can be improved.
Recall a saturated arc
f(u,v)=c(u,v) A saturated arc is full to its capacity.
Recall the spare capacity of an arc
spare=c(u,v)−f(u,v) This is how much more flow the arc could still carry.
Recall a flow-augmenting path
a path from S to T with spare capacity throughout Along such a path the value of the flow can be increased.
Recall forward and backward arcs on an augmenting path
forward: c−f,backward: f Forward arcs use spare capacity; backward arcs release existing flow.
State the value of the flow
val(f)=8 This is the net flow out of the source.